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Melody  11 Feb 2022
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MathPong:

Quadratic Functions

I need help with a question i got on my math book. It goes like this Solve the equation y = 0 graphically and algebrally. And i got a formula for this and it goes like: y = x*h and the h in the formula is 12 - x so i did this x(12-x) and i got this 12x-x^2. But its very confusing please help me its a Quadratic Function..



Thanks Walt, I just want to talk about this question because there is an easier way to do it.
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To solve algebraically you have x(12-x)=0
now if you multiply 2 things together and get 0 for an answer it means that one or the other or both must equal 0. Just think about that fo a moment and convince yourself that it is true.
Now
If x(12-x)=0 then
either x=0 or 12-x=0 and if 12-x=0 then x=12
so the algebraic solution is x=0 and x=12
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It is a bit strange to be asked to solve this graphically but because you effectively use the algebraic solution to graph it anyway.
But i will talk about it.
y= -x^2 +12x is a concave down parabola. It's a parabola because the highest power of x is 2. and it is concave down because the co-efficient of x^2 is negative.
Now I am going to put it back into the previous form
y=x(12-x)
The roots are 0 and 12
The roots are where y=0, or to put it another way they are the x intercepts (where the curve crosses the x axis)
The vertex (if you want to find it and you really don't need to) will have an x=(0+12)/2 = 6 SO y=6(12-6)=36
That is, the vertex is (6,36)
Now you have 3 points, that is enough to define a parabola, so draw the parabola.
Graphically, to solution to
0 = x(12-x) is the intersection of y=0 (the xaxis) and y=x(12-x)
you can read the intersection points off your graph. x=0 and x=12.
20 Jan 2014
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guest:

Dear All,

Please help me about the Applications of Differentiation - Optimization below :

=============================================================

The demand function is : P = 400 - 2Q

Average Cost : AC = 0.2Q + 4 + (400/Q)

Questions :

a. Determine how many units of a product that must be produced if the company wants to get the maximum profit?
b. Determine the selling price when the profit is maximum.
c. Determine the maximum profit that can be earned by the company.



==============================================================

Answer :

P = 400 – 2Q

TR = P.Q = (400 – 2Q)Q = 400Q – 2Q2

TC = AC.Q = (0.2Q + 4 + (400/Q))Q = 0.2Q2 + 4Q + 400

Profit = TR – TC
= (400Q – 2Q2) – (0.2Q2 + 4Q + 400)
= – 2.2Q2 + 396Q – 400

I don't know the next step, please help me..

Thank you very much for your answer..



Ok, so far, so good.
a. Determine how many units of a product that must be produced if the company wants to get the maximum profit?
The profit function is a concave down parabola (parabola because the biggest power of Q is 2, and concave down because the coefficient of Q 2 is negative (-2.2))
The maximum point wil be when the gradient of the tangent is 0
That is when dProfit/dQ=0 [Note: Marginal profit = MProfit = dProfit/dQ]
So Mprofit = -4.4Q + 396 (I have just differentiated profit)
Solve Mprofit = 0 and you will get the quantity that will give the maximum profit.

b. Determine the selling price when the profit is maximum.
Substitute the Q that you just got into the P function to get the selling price per unit

c. Determine the maximum profit that can be earned by the company.
substitute the Q that you just got into the Profit function.
20 Jan 2014